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Reptile [31]
2 years ago
10

In t-ball, young players use a bat to hit a stationary ball off a stand. The 140 g ball has about the same mass as a baseball, b

ut it is larger and softer. In one hit, the ball leaves the bat at 12 m/s after being in contact with the bat for 2.0 ms . Assume constant acceleration during the hit.
Physics
1 answer:
fomenos2 years ago
3 0

Question is missing. Found on google:

<em>"Part A What is the acceleration of the ball? Express your answer to two significant figures and include the appropriate units. </em>

<em>Part B </em>

<em>What is the net force on the ball during the hit?  </em>

<em>Express your answer to two significant figures and include the appropriate units."</em>

Solution:

A) 6000 m/s^2

The acceleration of the ball is given by

a=\frac{v-u}{t}

where

v = 12 m/s is the final velocity

u = 0 is the initial velocity (the ball is stationary)

t = 2.0 ms = 0.002 s is the time of contact

Substituting,

a=\frac{12-0}{0.002}=6.0 \cdot 10^3 m/s^2

B) 8.4\cdot 10^2 N

The force on the ball can be found by using Newton's second law:

F=ma

where

m = 140 g = 0.14 kg is the mass of the ball

a=6.0\cdot 10^3 m/s^2 is the acceleration

Substituting,

F=(0.14)(6.0\cdot 10^3)=8.4\cdot 10^2 N

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A 2 ft x 2 ft x 2 ft box weighs 100 pounds, and the weight is evenly distributed. What is the magnitude of the minimum horizonta
Tems11 [23]

Answer:

Explanation:

Let the force required be F . It is applied at the top of the box . The box is likely to turn about a corner . Torque of this force about this corner

= F x 2

This torque will try to turn the box . On the other hand the weight which is acting at CM will create a torque about the same corner . This torque will try to prevent the box to turn around the corner.

This torque of weight

= 100 x 1

= 100 pound ft.

For equilibrium

Torque of F = torque of weight.

F x 2  = 100

F = 50 pounds .

7 0
2 years ago
Read 2 more answers
luggage handler pulls a 20.0 kg suitcase up a ramp inclined at 34.0 ∘ above the horizontal by a force F⃗ of magnitude 165 N that
Andreas93 [3]

Answer:

a)  W = 643.5 J, b) W = -427.4 J  

Explanation:

a) Work is defined by

       W = F. x = F x cos θ

in this case they ask us for the work done by the external force F = 165 N parallel to the ramp, therefore the angle between this force and the displacement is zero

       W = F x

let's calculate

       W = 165  3.9

        W = 643.5 J

b) the work of the gravitational force, which is the weight of the body, in ramp problems the coordinate system is one axis parallel to the plane and the other perpendicular, let's use trigonometry to decompose the weight in these two axes

       sin θ = Wₓ / W

       cos θ = Wy / W

        Wₓ = W sinθ = mg sin θ

        Wy = W cos θ

the work carried out by each of these components is even Wₓ, it has to be antiparallel to the displacement, so the angle is zero

      W = Wₓ x cos 180

      W = - mg sin 34  x

     

let's calculate

       W = -20 9.8 sin 34 3.9

        W = -427.4 J

The work done by the component perpendicular to the plane is ero because the angle between the displacement and the weight component is 90º, so the cosine is zero.

3 0
2 years ago
An object travels 50 m in 4 s. It had no initial velocity and experiences constant acceleration. What is the magnitude of the ac
Allisa [31]

Answer:

The formula to calculate velocity in this case:

v = v0 + at

=> a = (v - v0)/t

       = (50 - 0)/4

       = 50/4 = 12.5 (m/s2)

Hope this helps!

:)

3 0
2 years ago
Utility poles are to be set every 30 meters. How many poles will be set in one mile if one was set at the beginning of the mile)
Zepler [3.9K]

Answer:

53

Explanation:

Because there are 1609.34 meters in a mile. 1609.34÷30=53.64 but because you put one at the beginning of the mile it will stay 53 and not round up to 54

4 0
2 years ago
Read 2 more answers
The engine on a fighter airplane can exert a force of 105,840 N (24,000 pounds). The take-off mass of the plane is 16,875 kg. (I
FrozenT [24]

Answer:

The acceleration you can get with that engine in your car is around 70,56 (\frac{m}{s^{2} }) or 7,26 (\frac{ft}{s^{2} } ) using 1500kg of mass or 3306 pounds

Explanation:

Using the equation of the force that is:

F=m*a

So, you notice that you know the force that give the engine, so changing the equation and using a mass of a car in 1500 kg or 3306 pounds

a=\frac{F}{m} =\frac{105840 N }{1500 (kg) }

a=\frac{105840 (\frac{kg*m}{s^{2} } )} {1500 kg }

<em> Note: N or Newton units are: \frac{kg * m}{s^{2} }</em>

a= 70,54 \frac{m}{s^{2} }

Also in pounds you can compared

a= \frac{2400  lf }{3 306  lf}

Note: lf in force units are: \frac{lf*ft}{s^{2} }

a=7,26 \frac{ft}{s^{2} }

8 0
2 years ago
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